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<h1 class="heading"><a href="MATH-2023-OPDE.html"><span class="title">MATH 2023: Ordinary and Partial Differential Equations</span></a></h1>
<p class="byline">Xiaoyi Chen and Wei Zhang</p>
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<a href="ch_first.html" data-scroll="ch_first" class="internal"><span class="codenumber">1</span> <span class="title">Introduction</span></a><ul>
<li><a href="sec_1-intro.html" data-scroll="sec_1-intro" class="internal">Classification of Differential Equations</a></li>
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<a href="ch_second.html" data-scroll="ch_second" class="internal"><span class="codenumber">2</span> <span class="title">First Order Ordinary Differential Equations</span></a><ul>
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<a href="ch_seven.html" data-scroll="ch_seven" class="internal"><span class="codenumber">7</span> <span class="title">Partial Differential Equations</span></a><ul>
<li><a href="sec7_1.html" data-scroll="sec7_1" class="internal">Two-Point Boundary Value Problems</a></li>
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<li><a href="sec7_5.html" data-scroll="sec7_5" class="internal">Even and Odd Functions</a></li>
<li><a href="sec7_6.html" data-scroll="sec7_6" class="internal">Introduction to Partial Differential Equations</a></li>
<li><a href="sec7_7.html" data-scroll="sec7_7" class="internal">1D Heat Equation; Solutions by Separation of Variable and Fourier Series</a></li>
<li><a href="sec7_8.html" data-scroll="sec7_8" class="internal">Other Heat Conduction Problems</a></li>
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<main class="main"><div id="content" class="pretext-content"><section class="section" id="sec6_4"><h2 class="heading hide-type">
<span class="type">Section</span> <span class="codenumber">6.4</span> <span class="title">Repeated Eigenvalues</span>
</h2>
<p id="p-271">Consider</p>
<div class="displaymath process-math" data-contains-math-knowls="" id="eq7_13_1">
\begin{equation}
{\bf x}^{\prime}={\bf A}\,{\bf x}, \quad {\bf A} \mathrm{~is ~real}.\tag{6.4.1}
\end{equation}
</div>
<p class="continuation">The corresponding eigenvalue problem is</p>
<div class="displaymath process-math" data-contains-math-knowls="" id="eq7_13_2">
\begin{equation}
({\bf A}-r{\bf I}) \vec{\xi}={\bf 0}.\tag{6.4.2}
\end{equation}
</div>
<p class="continuation">Suppose that there is a <span class="process-math">\(k-\)</span> repeated eigenvalues <span class="process-math">\(r=\rho\text{,}\)</span> i.e. <span class="process-math">\(\rho\)</span> is the <span class="process-math">\(k-\)</span> fold root of</p>
<div class="displaymath process-math" data-contains-math-knowls="" id="eq7_14">
\begin{equation}
|{\bf A}-r {\bf I}|=0.\tag{6.4.3}
\end{equation}
</div>
<p class="continuation">For the corresponding eigenvectors, there are two possibilities:(i) There are <span class="process-math">\(k-\)</span> linear independent eigenvectors <span class="process-math">\(\vec{\xi}^{(1)}, \vec{\xi}^{(2)}, \cdots \vec{\xi}^{(k)}\text{.}\)</span> In this case, we have <span class="process-math">\(k-\)</span> linear independent solutions</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{\bf x}^{(1)}=\vec{\xi}^{(1)} e^{\rho t}, \quad
{\bf x}^{(2)}=\vec{\xi}^{(2)} e^{\rho t}, \cdots
{\bf x}^{(k)}=\vec{\xi}^{(k)} e^{\rho t}.
\end{equation*}
</div>
<p class="continuation">(ii) There are less than <span class="process-math">\(k\)</span> linear independent eigenvectors. In this case, we have to find other solutions.</p>
<p id="p-272"><dfn class="terminology">Example 1</dfn> Find the general solution of the following equation</p>
<div class="displaymath process-math" data-contains-math-knowls="" id="eq7_13">
\begin{equation}
{\bf x}^{\prime}={\bf A} \, {\bf x}=\left(
\begin{array}{cc}
4 &amp; -2\\
8 &amp; -4
\end{array}
\right) {\bf x}.\tag{6.4.4}
\end{equation}
</div>
<p id="p-273"><dfn class="terminology">Solution</dfn> Consider</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_11.html" id="eq7_11">
\begin{equation}
({\bf A}-r{\bf I}) \vec{\xi}={\bf 0}.\tag{6.4.5}
\end{equation}
</div>
<p class="continuation">We require</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_11.html">
\begin{equation*}
|{\bf A}-r {\bf I}|=0 \to 
\left|
\begin{array}{cc}
4-r &amp; -2\\
8 &amp; -4-r
\end{array}
\right|=0 \to r^2=0 \to r_1=r_2=0.
\end{equation*}
</div>
<p class="continuation">For <span class="process-math">\(r_1=r_2=0\text{,}\)</span> from (<a href="" class="xref" data-knowl="./knowl/eq7_11.html" title="Equation 6.4.5">(6.4.5)</a>),</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_11.html" id="eq7_12_1">
\begin{equation}
\left(
\begin{array}{cc}
4 &amp; -2\\
8 &amp; -4
\end{array}
\right) 
\left(
\begin{array}{c}
\xi^{(1)}_1\\
\xi^{(1)}_2
\end{array}
\right)={\bf 0} \to 
\begin{array}{c}
4 \xi^{(1)}_1-2 \xi^{(1)}_2=0\\
8 \xi^{(1)}_1-4\xi^{(1)}_2=0
\end{array}
\to \xi^{(1)}_2=2 \xi^{(1)}_1.\tag{6.4.6}
\end{equation}
</div>
<p class="continuation">Choose <span class="process-math">\(\xi^{(1)}_1=1\text{,}\)</span> the eigenvector is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_11.html">
\begin{equation*}
\vec{\xi}^{(1)}=\left(
\begin{array}{c}
1 \\
2
\end{array}
\right)
\end{equation*}
</div>
<p class="continuation">and one solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_11.html">
\begin{equation*}
{\bf x}^{(1)}=\vec{\xi}^{(1)} e^{r_1 t}=\left(
\begin{array}{c}
1 \\
2
\end{array}
\right).
\end{equation*}
</div>
<p id="p-274">Seek another solution of the form</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_12.html ./knowl/eq7_13.html" id="eq7_12">
\begin{equation}
{\bf x}^{(2)}=t \vec{\xi} e^{r_1 t}=t \vec{\xi},\tag{6.4.7}
\end{equation}
</div>
<p class="continuation">where <span class="process-math">\(\vec{\xi}\)</span> is to be determined. Substituting (<a href="" class="xref" data-knowl="./knowl/eq7_12.html" title="Equation 6.4.7">(6.4.7)</a>) into (<a href="" class="xref" data-knowl="./knowl/eq7_13.html" title="Equation 6.4.4">(6.4.4)</a>), we have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_12.html ./knowl/eq7_13.html">
\begin{equation*}
\vec{\xi}=\left(
\begin{array}{cc}
4 &amp; -2\\
8 &amp; -4
\end{array}
\right) t \vec{\xi}.
\end{equation*}
</div>
<p class="continuation">The left hand side is irrelevant with <span class="process-math">\(t\)</span> while the right hand side is and this can not be satisfied except <span class="process-math">\(\vec{\xi}={\bf 0}\text{.}\)</span></p>
<p id="p-275">Try a solution of the form</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_13.html ./knowl/eq7_12_1.html">
\begin{equation*}
{\bf x}^{(2)}=(t \vec{\xi}+\vec{\eta}) e^{r_1 t}=t \vec{\xi}+\vec{\eta},
\end{equation*}
</div>
<p class="continuation">where <span class="process-math">\(\vec{\xi}, \vec{\eta}\)</span> are to be determined. Substituting it into (<a href="" class="xref" data-knowl="./knowl/eq7_13.html" title="Equation 6.4.4">(6.4.4)</a>):</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_13.html ./knowl/eq7_12_1.html">
\begin{equation*}
\vec{\xi}={\bf A} (t\vec{\xi}+\vec{\eta})=t {\bf A} \, \vec{\xi}+{\bf A}\, \vec{\eta}.
\end{equation*}
</div>
<p class="continuation">Compare both sides:</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_13.html ./knowl/eq7_12_1.html" id="eq7_15">
\begin{equation}
{\bf A} \, \vec{\xi}={\bf 0},\quad
{\bf A} \, \vec{\eta}=\vec{\xi}.\tag{6.4.8}
\end{equation}
</div>
<p class="continuation"><span class="process-math">\((\ref{eq7_15})_1\)</span> is the same as (<a href="" class="xref" data-knowl="./knowl/eq7_12_1.html" title="Equation 6.4.6">(6.4.6)</a>), thus</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_13.html ./knowl/eq7_12_1.html">
\begin{equation*}
\vec{\xi}=\vec{\xi}^{(1)}=\left(
\begin{array}{c}
1\\
2
\end{array}
\right).
\end{equation*}
</div>
<p class="continuation">From <span class="process-math">\((\ref{eq7_15})_2\text{,}\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_13.html ./knowl/eq7_12_1.html">
\begin{equation*}
\left(
\begin{array}{cc}
4 &amp; -2\\
8 &amp; -4
\end{array}
\right) \left(
\begin{array}{c}
\eta_1\\
\eta_2
\end{array}
\right)=\left(
\begin{array}{c}
1\\
2
\end{array}
\right) \to \eta_2=2 \eta_1-1/2.
\end{equation*}
</div>
<p class="continuation">Choose <span class="process-math">\(\eta_1=0\text{,}\)</span> we have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_13.html ./knowl/eq7_12_1.html">
\begin{equation*}
\vec{\eta}=\left(
\begin{array}{c}
0\\
-1/2
\end{array}
\right).
\end{equation*}
</div>
<p id="p-276">Then, the second solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{\bf x}^{(2)}=t \vec{\xi}+\vec{\eta}=t \left(
\begin{array}{c}
1\\
2
\end{array}
\right)+\left(
\begin{array}{c}
0\\
-1/2
\end{array}
\right).
\end{equation*}
</div>
<p class="continuation">It can be shown that <span class="process-math">\(W[{\bf x}^{(1)}, {\bf x}^{(2)}] \neq 0\text{,}\)</span> thus the general solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{\bf x}=C_1  \left(
\begin{array}{c}
1\\
2
\end{array}
\right)+C_2 \left[t \left(
\begin{array}{c}
1\\
2
\end{array}
\right)+\left(
\begin{array}{c}
0\\
-1/2
\end{array}
\right) \right]
\end{equation*}
</div>
<p id="p-277"><dfn class="terminology">Remark</dfn> Consider the situation where <span class="process-math">\(r=\rho\)</span> is a triple-repeated eigenvalue of (<a href="" class="xref" data-knowl="./knowl/eq7_14.html" title="Equation 6.4.3">(6.4.3)</a>):<dfn class="terminology">Case i.</dfn> There is only one eigenvector <span class="process-math">\(\vec{\xi}^{(1)}\)</span> and one solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_14.html ./knowl/eq7_13_1.html ./knowl/eq7_13_1.html">
\begin{equation*}
{\bf x}^{(1)}=\vec{\xi}^{(1)} e^{\rho t}.
\end{equation*}
</div>
<p class="continuation">The second and third solutions have the form of</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_14.html ./knowl/eq7_13_1.html ./knowl/eq7_13_1.html">
\begin{equation*}
{\bf x}^{(2)}=(t \vec{\xi}^{(2)}+\vec{\eta}^{(2)}) e^{\rho t},\quad
{\bf x}^{(3)}=(t^2 \vec{\xi}^{(3)}+t \vec{\eta}^{(3)}+\vec{\zeta}^{(3)})  e^{\rho t},
\end{equation*}
</div>
<p class="continuation">where <span class="process-math">\(\vec{\xi}^{(2)}, \vec{\eta}^{(2)},  \vec{\xi}^{(3)}, \vec{\eta}^{(3)}, \vec{\zeta}^{(3)}\)</span> can be determined from (<a href="" class="xref" data-knowl="./knowl/eq7_13_1.html" title="Equation 6.4.1">(6.4.1)</a>).<dfn class="terminology">Case ii</dfn> There are two linear independent eigenvectors <span class="process-math">\(\vec{\xi}^{(1)}\)</span> and <span class="process-math">\(\vec{\xi}^{(2)}\text{.}\)</span> Two solutions are</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_14.html ./knowl/eq7_13_1.html ./knowl/eq7_13_1.html">
\begin{equation*}
{\bf x}^{(1)}=\vec{\xi}^{(1)} e^{\rho t},\quad
{\bf x}^{(2)}=\vec{\xi}^{(2)} e^{\rho t}.
\end{equation*}
</div>
<p class="continuation">The third solution has the form</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_14.html ./knowl/eq7_13_1.html ./knowl/eq7_13_1.html">
\begin{equation*}
{\bf x}^{(3)}=(t \vec{\xi}^{(3)}+\vec{\eta}^{(3)}) e^{\rho t}
\end{equation*}
</div>
<p class="continuation">and by taking it into the (<a href="" class="xref" data-knowl="./knowl/eq7_13_1.html" title="Equation 6.4.1">(6.4.1)</a>), we have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_14.html ./knowl/eq7_13_1.html ./knowl/eq7_13_1.html" id="eq7_16">
\begin{equation}
({\bf A}-\rho {\bf I}) \vec{\xi}^{(3)}=0,\quad
({\bf A}-\rho {\bf I}) \vec{\eta}^{(3)}=\vec{\xi}^{(3)},\tag{6.4.9}
\end{equation}
</div>
<p class="continuation">where <span class="process-math">\(\vec{\xi}^{(3)}\)</span> should be one eigenvector corresponding to <span class="process-math">\(r=\rho\text{.}\)</span> Most generally, it should be</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq7_14.html ./knowl/eq7_13_1.html ./knowl/eq7_13_1.html">
\begin{equation*}
\vec{\xi}^{(3)}=k_1 \vec{\xi}^{(1)}+k_2 \vec{\xi}^{(2)}
\end{equation*}
</div>
<p class="continuation">where we should choose such values for <span class="process-math">\(k_1\)</span> and <span class="process-math">\(k_2\)</span> that <span class="process-math">\((\ref{eq7_16})_2\)</span> has a solution for <span class="process-math">\(\vec{\eta}^{(3)}\text{.}\)</span></p></section></div></main>
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